Aptitude - Permutation and Combination - Discussion

Logarithms ===> total = 10 alphabets.

We have to form 4-letter words, without repeating the letters. We have place for 4 alphabets.

In 1st blank, out of 10, anyone of it can come, so we have 10 choices ---> 10.

In 2nd blank, now, 9 letters are left, out of 9 you can choose one ----> 10*9.

In 3rd blank, now 8 left, i.e. ---> 10*9*8.

In 4th blank, any remaining of 7 letters can come ---> 10*9*8*7 = 5040.

Here we are using --- Logarithms --- word which contain different alphabets.

If it contain one or two alphabets more then one time then above solution is applicable.

Solve this example in place of Logarithms use logarithm. Please solve this.

Firstly we have to check if any alphabet is repeated in the word "LOGARITHMS".

No, alphabets is repeated so it can be treated as any other general case. We are using permutation because it is not just the selection of 4 alphabets from the given word but the rearrangement of these alphabets in order to form a word (given: The words may or may not have any meaning). Now there are 10 alphabets & we must find out the number of 4 lettered words that can be formed which can be done in 10p4 ways. We need not multiply with 4! because have already considered the rearrangements (i.e usage of the concept of permutations).

The given word consists of 10 letters.

No letters are repeated
we have to form a four-letter word and letters should not be repeated so;
_ _ _ _
Consider 4th place we have now 10 letters and one place to fill, So 10C1 possibilities are there
Now consider 3rd place we have 9 letters(one already used)and one place to fill, so 9C1 possibilities are there.

Now consider 2nd place we have 8 letters and one place to fill, so 8C1 possibilities are there
similarly, for 1st place, 7C1 possibilities are there.

Now multiplying them =10C1x9C1x8C1x7C1=10x9x8x7=5040.

Logarithms===> total= 10 alphabets.. okay
we have to form 4-letter words, without repeating the letters.
_ _ _ _
we have place for 4 alphabets
In 1st blank, out of 10, any1 of it can come,so we have 10 choices--->10
In 2nd blank, now,9 letters are left, out of 9 u can choose one---->10*9
In 3rd blank, now 8 left, i.e.--->10*9*8
In 4th blank, any remaining of 7 letters can come---> 10*9*8*7
= 5040

Easier method.

Here, repetition is not allowed. So, to form a word of 4 letters. _ _ _ _.

The 1st letter can be chosen out of 10 letters.

The 2nd letter can be chosen out of the remaining 9 letters. (since no repetition & 1 letter is filled in the 1st position).

The 3rd letter can be chosen out of the remaining 8 letters.

The 4th letter can be chosen out of the remaining 7 letters.

Thus, 10 * 9 * 8 * 7 = 5040.

We use permutation for selection and arranging.
They asked two processes in question.

step 1: Selection- to select 4 letters from 10 letters,
step 2: Arranging- to arrange 4 letters in different ways.

Why use arranging?

Because, by arranging the 4 words in different ways we can get many words, which can be meaningful or non-meaningful.
So, for permutation = n!/(n-1)!

Its very easy, no need to get confused with permutation.

Divide the statement in two part.

From given 10 character, we have to choose 4 of them only.
So, 10C4 = 10*9*8*7/4! = 210.

Now we have to arrange there 4 character.
So, it would be simply 4! because there is no repetition of word.

Total - 210*24 = 5040.

Moral - choose 4 word out of 10 then arrange them by 4!.

If the letters can be repeated then the answer would be 10^4.

Because if the word is _ _ _ _.

The 1st blank can be filled in 10 ways.

The 2nd blank can be filled in 10 ways again (since letters can be repeated).

Similarly, 3rd and 4th blanks can also be filled in 10 ways.

So, according to the fundamental principle the number of ways =10*10*10*10=10^4.

The formula we hv usd here is = n!/(P1!)(P2!)...(Pr!)
Here n=10(no. of letters) and P1,P2.. are the different letters of the word.
Then the no. of permutation of these 10 letters is
= 10!/(1!)(1!)..upto 10th letter [(1!)...upto 10th letter used bcoz each letter occured only once]
= 10*9*8*7 [bcoz we hv to make 4 letter words]
= 5040

When we form LOGA and AGOL it is permutation but in combination we don't write like r, s is similar to s, r -in combination, not in permutation. So in above question we are not allowed to double the letter but are allowed to use the same letter in different words. .